Giới thiệu phương pháp và kĩ thuật ôn nhanh thi đại học đạt điểm cao môn Toán: Phần 1

Phương pháp và kĩ thuật ôn nhanh thi đại học đạt điểm cao môn Toán cuốn sách hay dùng để ôn tập những kiến thức môn toán để đạt điểm cao. Bên cạnh đó còn hướng dẫn các em cách giải bài tập toán nhanh và hiệu quả, nhằm mục đích giúp các em đạt điểm cao nhất trong các bài kiểm tra, kì thi sắp tới.

PH561P ncuYin P I I U Kiirinii

PHUONO PHAP

D A T D I E M C A O

NGUYEN PHU KHANH

PHUONG PHAP

&

KI T H U A T

DAT DIEM CAO

\SP/ NHA XUAT BAN DAI HOC SlT PHAM

hoc sinh on tap, van dung sang tao kien thuc, tong hop, phan tich, kiem tra nang lye toan dien, hucfng dan chien thuat giai de thi dai hoc Sach trinh bay noi dung trong tam on thi dai hoc gbm:

- Phucmg trinh, bat phuong trinh, he phucmg trinh dai so He phucmg trinh mil va Idgarit

- Bai todn to'ng hap Bat dang thuc; cue tri cua bieu thuc dai so

- 8 phuang phdp giai phuang trinh lugng gidc nhanh nhat

- Van de lien quan den sophiec, dai sotohffp, xdc sudi

- Cdc bai todn lien quan den ung dung ciia dqo ham va do thi cua ham so: chieu bieh thien cua ham so; cue tri; gid tri Ion nhat va nho nhat cua ham so; tiep tuyen, tiem can (dimg vd ngang) cua do thi ham so; tint tren do thi nhirng diem c6 tinh chat cho truac, luang giao giua hai do thi (mot trong hai do thi Id dubng thang)

- Tim giai han

- Tim nguyen ham, tinh tich phan Ung dung cua tich phan: Tinh dien tich hinh phang, the tich khdi trdn xoay

- Hinh hoc khong gian (tong hap): quan he song song, quan he vuong goc cua dubng thang, mat phdng; dien tich xung quanh cua hinh non trdn xoay, hinh tru trdn xoay; the tich khdi lang tru, khdi chop, khdi non trdn xoay, khdi tru trdn xoay; tinh dien tich mat can vd the tich khdi cdu

- Phuang phdp toa do trong mat phang: Xdc dinh toa do cua diem, vecta Duong trdn, ba dudng conic Viet phuang trinh dudng thdng Tinh gdc; tinh khodng cdch tit diem den dudng thang

- Phuang phdp toa do trong khdng gian: Xdc dinh toa do cua diem, vecta Duong trdn, milt cdu Viet phuang trinh mat phdng, dudng thdng Tinh goc; tinh khoang cdch tie diem den dudng thang, mat phdng; khodng cdch giita hai dubng thdng; vi tri tuang ddi cua dubng thdng, mat phdng vd mat cdu

Mac du tac gia da danh nhieu tam huyet cho cuon sach, nhung sai sot la dieu kho tranh khoi, rat mong nhan duac sy phan bien va gop y quy bau ciia ban doc

de nhung Ian tai ban sau cuon sach dugc hoan thien hon

Tac gia

3

C A U T R U C D E T H I DAi H Q C M O N T O A N '

I PHAN C H U N G (7 diim)

Cau 1 (2 diem):

a) Khao sat su bien thien va ve do thi cua ham so

b) Cac bai toan lien quan den ung dung cua dao ham va do thi cua ham so: chieu bien thien ciia ham so; cue tri; gia trj idn nhat va nho nhat cua ham so; tiep tuyen, tiem can (dung va ngang) cua do thi ham so; tim tren do thi nhiing diem c6 tinh chat cho truac, tuong giao giira hai do thi (mot trong hai do thi la duong thang)

- Tim gioi han

- Tim nguyen ham, tinh ti'ch phan

- ung dung ciia tich phan: tinh dien tich hinh phang, the tich khoi tron xoay

Cau 5 (1 diim):

Hinh hoc khong gian (tong hop): quan he song song, quan he vuong goc cua duong thang, mat phang; dien tich xung quanh cua hinh non tron xoay, hinh tru tron xoay; the ti'ch khoi lang tru, khoi chop, khoi non tron xoay, khoi tru tron xoay; tinh dien tich mat cau va the tich khoi cau

Cau 6 (1 diem):

Bai toan tong hop

II PHAN R I E N G (3 diem)

Thi sink chi dugc lam mot trong hai phan (phan A hoac phan B)

A Theo chucmg trinh Chuan:

Cau 7a (1 diem):

Phuong phap toa do trong mat phang:

- Xac dinh tpa dp cua diem, vecto

- Duong tron, elip

- Viet phuong trinh duong thSng

- Tinh goc; tinh khoang each t u diem den duong thing

Cau 8a (1 diim):

Phuong phap toa do trong khong gian:

- Xac djnh toa dp cua diem, vecto

- Duong tron, mat cau

- Tinh goc; tinh khoang each t u diem den duong thang, mat phang; khoang each giiia hai duong thang; vi tri tuong doi cua duong thang, mat phang va mat cau

Cau 9a (

1 diem)

:

- S

o phuc

- T

o hop, xa

c suat, thon

g ke

B Theo chuang trinh

Nang cao:

Cau 7b

- Duon

g tron, b

a duon

g conic

Cau 8b

- Duon

g tron, ma

t cau

g thang

t phan

g v

a ma

t cau

Cau 9b

(1 diem)

:

- S

o phuc

- T

o hop, xa

c suat, thon

g ke

- Ba

t dan

g thiie Cu

PHlTdNG TRINH , BA

T PHlTdN

G TRINH , H

E

I

P HlT

O fNG TRINH

Phuong trinh, bat phuong

trinh, he phuong trinh

dai so.

H( phuong trinh mil

G T RI NH C

3) He doi xung loai 1

Phuong phap chung: Dat an phu a = x + y; b = xy

4) He doi xung loai 2

Phuong phap chung: Tru tung vehai phuong trinh da cho nhau ta dugc: (x - y).f(x; y) = 0

5) He phuong trinh dang cap bac hai

Xet truong hop y = 0 Vol y ?t 0, ta c6 the tien hanh theo cac each sau:

- Dat an phu y = t.x

- Chia ca hai ve'cho y^, va dat t = —

y

M O T SO P H U O N G P H A P G I A I H E P H U O N G T R I N H

1) Phuong phap the

• Phuong phap: Ta riit mot an (hay mot bieu thuc) t u mot phuong trinh trong he va

the vao phuong trinh con lai

• Nhqn dang: Phuong phap nay thuong hay su dung khi trong he c6 mot phuong

trinh la bac nha't doi voi mot an nao do

2) Phuong phap cong dai so

3) Phuong phap bien doi thanh tich

4) Phuong phap dat an phu

5) Phuong phap ham so

6) Phuong phap su dung bat dang thuc

C/iM y: ting dung dao ham gidi todn phuong trinh, he phuong trinh

Su dung cac tinh chat ciia ham so de giai phuong trinh la dang toan kha quen thupc

Ta thuong c6 ba huong ap dung sau day:

Huang 1: Thuc hien theo cac buoc:

Buac 1: Chuyen phuong trinh ve dang: f(x) = k

Buac 2: Xet ham so y = f(x)

Buac 3: Nhan xet:

• Voi x = Xg <=> f(x) - f(xg) = k, do do Xg la nghiem

• Voi X > Xg o f(x) > f(xQ) = k, do do phuong trinh v6 nghiem

• Voi X < Xg <=> f(x) < f(xQ) = k, do do phuong trinh v6 nghiem

• Vay Xg la nghiem duy nhat aia phuong trinh

Huong 2: Thuc hien theo cac buoc:

Suae 1: Chuyen phuong trinh ve dang: f(x) = g(x)

Buoc 2: Dung lap luan khang dinh rang f(x) va g(x) c6 nhung tinh chat trai ngugc

nhau va xac dinh Xg sao cho f ( X g ) = g(xg)

Buac 3: Vay Xg la nghiem duy nhat ciia phuong trinh

Buac 3:

Khi d

o f(u) = f(v) o

u = v

1 V5 x

-l

-7 3x

= 4

3 ~ + X = x

+ X XV2x

4 2'"'+2^''-9.2'''+18

3 = 2^''*'+2''"+9.2^*"

Lcri giai

1 Die

u kien: x > —

1

V3x +13)(

%/3x +

13 =

0 o

V sx -l = V sx TlS

Trucrng hap

2: V

Sx -l + V3x

+ 1

3 =

6 (1)

2, Die

u kien: 1 +

- ^

0 o > 0

4x

+ 3

+ Vd

i X > 0 , phuon

g trin

h

o>

yx 2

+2

x =-2

(x 2

>0, t

a c

6 2t

^ +1-

6 Vx

^ +2

x =

1 o x^ + 2x =

a ma

n die

u kien

+ Vd

i X ^ -

2, phuon

g trin

h o

-V x^

+2x = -2(x

^ + 2x) +

3

Dat t = >

/x

^+

2x,t>

0, t

a c6: 2t

^ 1-

-3 =

0 o

t = | hoa

9 <=

> 4x

^ + 8x -

'^'^''^ man

dieu kien

Vay phuong trinh da cho c6 nghiem x = - 4 - V 5 2 ; X = - 1 + r/2

Phuong trinh da cho c6 nghiem duy nhat x = 2

4 Phuong trinh da cho tuong duong v6i

gidi

1 Die

u kien: x > 1;

x # 3;x #

15

= ' 0g 5^

log5|x-5

o2

|x-3 = x-l<:=>2x-

6 =

-l hoa

c 2x-6 = l-x <:=>x =

6 <

x <

4 v

a x ;t -

^ {

x + 6)

o log

^ (4 ~ x) + iog^ (

x + 6) =

1 + log^ |

x + 2|

o

(4 x)(x + 6) = 4|x +

2

<=>4(x + 2) = (4-x)(x + 6) hoa

c 4(

x + 2)

(4-x)(

=-x + 6) (v

i (*

) ne

n (4-x)

6x-1

6 = 0c:>x = 2,

x = -

8

+ V6

i x'-2x-32 = 0<=>

x =

l + 733,x = l-V33

; x = 1

- V3

3

3 Die

u kien: 2.6"

-4

" = 2".2"^2.6

" -4

" <

^2" (4

" + 2.6

" (4.6

" -3.4") <

^4" (2

" + V4.6

"-3.4")

= 8" +i2

27"-12"+2.8" = 27" +8

Vi d

u 3 Gia

1.2x^+6x^+6x + l =

3| x

+ 2

2 8

log4 N/X^ -

9 +

3^2 log4

(x + 3)^ =

^i

±i =

y 3+

3y2+

3y + iox = 2y^

+6y^

+6

y (1)

Theo de bai ta CO p h u a n g t r i n h 2x + 6 x + 6 x + l = y + 1 <=> y = 2x + 6 x + 6 x ( 2 ) ,

Ttr (1) va (2) s u y ra

o (x - y)(2x2 + 2y2 + 2xy + 6x + 6y + 7) = 0

2 Die

u kien: x > 1

Dat a = Vx-l,b = Vx

\/(

x-l)(x

X +

1 = a.b

:i

x-l = l<=>

x =

2

+ V6

i b = l th

i Vx^+x^+

x +

1 =l»

x''+x

^+

x = Oc=>x^x^+x +

j = 0<=>x = 0(loai,do

moi x)

l

=0

3 4(

N/X +T

-3

)X

' +

(I 3V

X +

1-8

)X

~4

VX

-1

3

-0

1 Die

u kien: -

2 <

x <

2

Dat t = x + \/4-x' =

^c

>3t'-2t-8 = 0ci>

t = hoac t

= 2

+ Vo

i t = 2, t

a c6: x + V4-x' =

a C

O x

+v4

-x^ = — <

= > v4

-x^ =

x>

2-0

4~x^ =4-4

x + x^

-i

3

x = -2±Vl4

=> X =

3

-2-Vl4

; x = 2

; x -2-Vi4

2 Die

u kien: x > -

l =

0

3 Dieu kien: x>l Bien doi phuong trinh da cho, ta dugc:

4 X ' N / > O ^ - 1 2 X ' + 1 3 x > / ) r r i - 8 x - 4 V x ^ l - 3 = 0

x V ^ ( 2 V ^ - 3 ) ' = 0 5

De vetrai bang 0 thi ta phai c6: <=> x =•

Vay X = ^ la nghiem cua phuong trinh da cho

' 2 3 ' 2 3

2 x ' * + 2 x ^ y + x^y^ = 2 x + 9(l) X ' + 2 x y = 6x + 6 (2)

Trumig hap

1: x = 0 khong tho

a ma

n (2)

Trudng hap 2: x^O,

(2) su

y r

a y = 6x + 6-x

2x , the'va

o (1) t

a dug

c ,

( 2 ^

6x + 6-x

+ x'

( 2 ^

^) + ^^^"^^ ^

^ =2

x +

9 ox(

x + 4)^ =0ci>

b (a

;b

>0)

4

He tr

o thanh:

a + 6b

= n + b '

b +

4a =

7 + a~

< =>

\

a-2 = (b-3)- (1)

b-3 = (a-2)' (2)

Lay (l)-(

2) t

a dugc: (b-a-l)(a + b-4) =

0

+ V6

i b-a-l =

0 =

> b = a + l tha

y va

o (1) t

a dugc:

a =

2 =

> b = 3 =>

(x

;y) = (log, 2;log, 3) hoa

c a = 3 => b = 4 =>

- 4 = 0 v6 nghie

m (d

o t

u (1), (2) t

a ru

t dug

c a ^ 2

; b ^ 3

l;2)

1

2x^-9y^

=(

y)(4xy-l)

xx"^ 3xy+ y

y + 4) = -36

1

Lai giai

-9y3=(x

x" 3x

-y + y^

=-1

2x^ -9y^

=(

y)(4xy-l) J2x

x-^ -9y'' = (x y)(x^ +

xy + y^)

x2-3xy + y2=

-l '

^-x^ -3xy + y

^

=-1

x = 2y

< =>

x = 2;y =

l

x = -2;y = -1

: (2;1)

; (-2

; -1)

2

1 - =

(1)

y-(x-4y)(2x-

y + 4) = -36 (2)

1 1

/ X

(y-x)(y

^+

xy+ ,

- y^

Trumg hap 1: x = y , the vao phuong trinh (2) ta dugc

Khi do (2) o 2x^ + 4y^ ~ 9xy + 4x - 16y = -36 o 2(x +1)^ + 4(y - 2f - 9 x y = -18

Truong hop nay khong xay ra do xy < 0 => 2(x +1)^ + 4{y - 2)^ - 9xy > 0

Vay nghiem cua he phuong trinh la (x;y) = (2; 2); (-6;-6)

V i du 8 Giai cac he phuong trinh sau tren tap so thuc:

Truong hap 2: (x + y)(x + y + 4) - 2xy = 0 <r> x^ + y^ + 4(x + y) = 0, v6 nghiem do dieu kien

Vay nghiem cua he la (x; y) = (-3; 7); (2; 2)

2 Dieu kien: x^- ,y> -

Trucntg hop 1: y

-x = 0<=>y =

x th

e va

o (1 ) ta

a duo

c V2-t

^ = 2-1 <=

<=> t = 1

2-t

^O

2-\}

4t +

xy

2-U J2

y nha

t (x

; y) = (1

; 1)

n ta

p so thuc:

1 <

2x^ +

xy = y^ - 3y +

2

x2 -y 2

2c:.y^-(x + 3)y + 2-2x^ =

0

Coi da

y la phuong trin

c y

= -x

+ l

Truang hap 1:

Truang hap 2:

y = 2x + 2

Jy = 2x +

2

x2 -y

^=

3 [3

X^+8 X

+ 7 =

0(A

'

= 5<0 )'

y =

-x + l |y =

2

(xy + l)x^+(x +

l)^ =x^

y + 5

x (1 )

4x

^y

+ 7x2 + 2x2

^9+1

= 2

x +

1 (2)

Dieu kien : y > -1 v

Tu phuon

g trin

h (1 ) ta c6:

x2 (xy + 1) + (x

-\

-x

(x

y + 1) = 0 <=> x(x

y + l)(x - 1) + (x- 1)^

= 0

o(x -l) (x 2y

+

x -l)

= 0ox = l hoa c x2y +

x -l

= 0

+ Vo

i x = 1 tha

y va

o (2 )

ta duoc

y + l +1) =

0 o y = -1

+ Vo

i x^y + 2 x-l = 0 o

Phuong t r i n h (*)<=> x = y, thay vao p h u o n g t r i n h t h u hai ciia he, ta d u g c

x^ - 2x^ - 3 x 4 - 4 = 0 <=>x = l hoac x = ^ - ^ ^ ^ hgp voi dieu kien ta dugc

l + \/l7

x = 1, x =

2 Vay he da cho c6 hai nghiem (x; y) la: (1; l}; Lt.^^^^y^ ^

Vi du

xy(x + l)(y + l) =

72 y(

2x)-2x = 1

y-0

1

H§ phuong

18

(x2+

x)(y2+

y) =

72

Dat \ x^

+x = a, a

> 4

y^

+y = b,

2

a =

12, b = 6

4

Truong hap 2: Do

7

y(

2x)-2x = 1

a dug

c h

e (b-a)^

-a

^ oa

^ = -2ab<=>a =

0 hoa

c a =-2

b

+ Vd

i a = 0 => b = ±

3 =: >

y = -

4

+ V6

i a = -2b

^5b2 = 9ob = 4-,

a = —

^ hoa

c b = -4=,

a = -

^

3 u

- 1

; 4);

-

(-1 —

^;

-1 + A) (_

sl5

Vi du

2>

/x

^+

3y-Jy^

(y + 3)-

1 Neu X = 0 thi he da cho <=>

Lai gidi

l + y 2 = 0 , , he nay v6 nghiem

4 - x ^ x^ +3y = 4

y^+8x = 9

y =

-4 - x ^

+ 8 x = 9

<

x^

-8x^

+72x-6

5 =

0 4-x

^ y =

(x l)(x + 5)(x^

-4

x + 13) =

0

x^

y = -7

x^

+x-5

y-2(

ll x- y) +

v =

^ y- x;u

,v > 0

'u = 2;v =

l

u -V =

1

7v-2u^

+4v^

-y =2

ll x-

y =

4

-X + y = 1

ri 3^ 2'2

y

\

Dat t = ^, t

a C

O phuan

g trinh: (

2 + t2)2

=3t2

(l +

t)

c =>

t'*-3t^

x - 5

Dat z = -J^^ > 0 Ta CO he phuong trinh

z = x + 5

2z = x^ + X- 5 <=>

z^ = x + 5 x^ = 2 z - x + 5 <=>

z^ =x + 5 x^ - z ^ = 2 z - 2 x

z = X + 5

X = z

x + z + 2 = 0

Truong hop x + z + 2 = 0 khong xay ra vi x, z > 0

Voi X = z > 0 ta duoc x = ^ ^ ^ ^ , y = 2x = 1 + V2I

Vay nghiem cua he da cho da cho la: (x;y) = ( ;2x = 1 +

Vi du 14 Giai he phuong trinh sau tren tap so thuc:

x^y^ + 4x^y - 3xy^ + x^ + y^ = 12xy + 3x - 4y +1

(u + l ) ( v + l ) = 2

3 u - 2 v = 3 <=>

2 o (u + l ) ( 3 u - l ) = 4

V i du 15 Giai cac phuong trinh sau tren tap so thuc:

Lai giai

1 Dieu kien: x > 3 Dat log3(x-2) = t > 0 C5> x = 3 * + 2

9)

\

= 1(1)

Ham

so f(t) =

v9 , nghich bien tren

M ne

n phuon

g trin

h (1) c

Voi t =

=:

> X

= 2 +

s

2 Die

u kien: x > 0

-1 = 3x - |x^ +

x +

1 j

log3(

X +X +

1 logj 3

-x = 3x - ^x^ +

X +

1

<=> I logj

^x

^

+X

+ 1J+ x

^ +

x +

j = log3 3x + 3x (*

)

Xet ha

m s

o t(t) = logg t +1 tren (0;

+00) c

6 f'(t) = +1

> 0, V

f (3x) <=>x^+

x +

l = 3xo

x = l

Vi d

u 16 Gia

1.21og(^

> /x +

N/X j

= log4

X 2

logj^

l + x

-N/

x

=(

x)V3

l-x

Lcri giai

1 Die

u kien: x > 0 Phuon

\[x +

\/x j = log

^ \/

x

Dat t = log(

, = log4

>/x

N/^ + ^ =

6'(1)

V^

= 4*(2)

The (2) va

o (1) t

a duoc: 4

* +2

^ = 6' ci

>

v6

y

v6

Xet ha

m s

o f (t) =

a tha

y f(t) nghjc

h bie

n tre

n # , m

y va

o (2) ta c6

x =

16

2 Die

u kien: x >

= N/

SVx ->/3

\/x^

Vay phuong trinh da cho c6 nghiem x = 0, x = 1

V i du 17 Giai cac phuong trinh sau tren tap so thuc:

f ( x ) = - ^ - ^ - ^ "^^^"^^^ + , i > 0,Vx 6 (-2; 4) f(x)dong bien tren | - 2 ; 4 J Ma

Lap bang bien thien suy ra: f (x) > f (3) = 2 Vx G [2;4]

=> Phuong trinh f (x) = \/x-2 + N / 4 - X = 2 C6 nghiem duy nhat x = 3

V i du 18 Giai cac phuong trinh sau tren tap so thuc:

Xethamso f(t) =

'+

^t, t

a c6: f (t) =

'ln

t + ^ >

0,Vt

Ham f(t) don

g bie

n tre

n M

nen phuong

trinh (*

)

= f

2x

^ ,

i v , , n die a ma 2 tho = = > X — < — — = — o

-u kien

2 Dieu kien: x # -3, x

— 1

0 713 X

a c6: f (t) = 713' I

n 71

3 >

0, V

t e ^ ne

n ha

m f(t) don

g bie

n

tren , do do phuong

-^^

ox^

+3x-40 = 0

x =

5

Vi d

u 19

1.

(6x + l)log^(x +

l) + 6(x

-l)

lo g2

(x + l)-

7 =

0

JM _>

i|

L ^ 2x ' el '- ^ " 2 ế

-5 x-

1

3 log,„,,(2014-l''"^'+4'""!-4) =

Lai giai

1.

Dieu kien: x > -

1

Ta tha

y x =

khong phai l

a nghie

m cii

a phuan

g trinh

7 1

;-—;+

oo ) (

^4

; "

Xetx 6 ( - — ; + o o ) , thi V T ( * ) la ham so dong bien, V P ( * ) la ham so nghich bien

6

ma V T ( I ) = V P ( l ) nen tren (-~;+co) thi x = 1 la nghiem duy nhat thoa man ( * )

6 Vay phuang trinh da cho c6 3 nghiem:

2 f)ieu kien: x 1; x - Phuong trinh da cho viet lai: e^^""''^ - ^ ^ ^ = e^" ^— (*)

2 x - 5 x - 1 Xet ham so f(t) = e' - - , t a c 6 f (t) = e' > 0,Vt ^ 0, nen ham f(t) dong bien V t # 0

x - 1 2x^5 <=> x = 2 hoac X = 4

nc:>f(|2x-5|) = f ( | x - l | ) o

Vay phuang trinh da cho c6 2 nghiem: x = 2 , x = 4

3 Dat T = 2 0 1 4 ' ' " " ' + 4 " " ' - 5

Phuong trinh da cho tro thanh: log,,,,^ (T + 1) = 2014^ - 1

Xet ham de thay phuong trinh tren luon c6 nghiem T = 0 <=> 2014^^^'""^ + 4'""' = 5 Data=|sinx| ( a 6 [ 0 ; l ] ) 2014-'+4^^' = 5 (1)

Ta c6: 2014'' ^ 4= (2) Dau " = " xay ra khi a = 0

Dat f(a) = 4 ' + 4 ^ ' ( 3 ) f (a) = l n 4 4 ' - l n 4 4 ^ ' ^

Xet f'(a) = Oc=>

4^:

(4).Dat g(z) = — , ( z e [ 0 ; l ] )

Khi do xet g(z) trong moi khoang ^ 10; U 1 ^

In 4 In 4 (4) •» a = V l- a ' <=> a =

Tiep tuc xet f(a) trong moi khoang

0;-Giai phuong trinh sin x = 1 hoac sin x = 0 danh cho ban doc

f(a) = 5 khi a = 1 hoac a = 0

V i du 20 Giai cac he phuang trinh sau tren tap so thuc:

Phuong trinh (l)

< =>

(x

- i

f + 3(x -1) = (2

- + 3(2

- y) (3)

Xetham

so f(t) = t

^+

3t,te(0

;+

oc) '

Ta c6

; f (t) = 2t +

3 >

0,Vt e (0;

2

y 6 (O;+oo) ne

n (3) <

= > f (x -1) =

f (

2 y) <

-= > x ^ 1 = 2

a dugc:

X

(3

-y )-

2 hoa

c y = 1

+ V

ai y = 1 => X = 2 (khong

thoa man dieu kien)

+ V

oi y = -

2 =

> X = 5 (thoa man dieu kien)

Vay h

e phuan

g trin

h c

6 nghie

m duy nhat (x

;y) = (5;-2)

2.Tac6 >

/t

^l+t

t>

0 Vtê

-e

"" =21n(

y + ^/y-+l)

gy-e >'

=2in(x + Vx^

+l)

Tu

do su

y ra: e

" e"" +

2 ln(

x + Vx^ +1) = ế - e"^ +

2 ln(

y + ^y^

+1)

Xet ha

m s

o f(t) = é -e

"' + 2ln(t +

7t^

+l)

2

Ta C

O f'(t) = é +

' + > 0

B

o d

o f(t) don

a su

y r

a x = ỵ

Thay x = y vao h

e t

a dugc:

ê-e-" =21n(

x + \/x

^+

l)og(x) = é'-e"

21n(x + Vx^

''-+l) =

0 (*)

Taco g'(x) = é'+e

——

^2Ve

la mo

t nghie

m cu

a (*)

Vi d

u 21 Cia

2ỵl

p-=0 L

4x-

26x-42.^

Xet ham so f(t) = (t^ + l)t => f'(t) = 3t^ +1 > 0, Vt e # => f(t) dong bien tren

Thir lai x = 3, y = 3 thoa man he phuang trinh

V i du 22 Giai cac he phuong trinh sau tren tap so thuc:

1 ( 2 x 2 - 3 x + 4 ) ( 2 y 2 - 3 y + 4) = 18

X + y +xy - 7 x - 6 y +14 = 0 2

x" - y'* + 28x + 31 = 32x + 4y^ + 6y^

x^ +y^ + x y - 7 x - 6 y + 14 = 0

Lai gidi

1

(2x^-3x + 4)(2y^

-3

y + 4) =

18 (1)

x^+y

^+

xy-7x-6y+ 1

4 =

0 (2)

6y+ 1

4, t

e =

> f'(t) = 4t -

3, f'(t) =

0 t = - <

1 4

so f(t) don

g bien

Trumtg hap 1:

6, ke

t ho

p vo

i y ^ 1

^ f(y) >

f(1) =

3 =

> f(x).f(y) =

(2x^ 3x + 4)(2y^

- 3

y + 4) >

18

Trumtg hap

2: x = 2

he tr

a than

h 2y2-3y + l =

0

<=>

y -4

y +

4 =

0 [y

^2

y = l,y =

2

v6 nghiem

x'* y^ + 28x +

31 = 32x + 4y^ +

6y^

(l)

x^+y

^+

xy-7x-6y + 1

4 =

0 (2)

Phuong trin

h (l)

o

x'* -32

x = y

"

+4y-^ +6y

^ -28y-3

1

c:>x''-32x = (

y +

32(y + l) (3)

f-Gia s

u phuan

g trin

h (2) l

7)

y-x +

6y + 1

^-4 =

0 c

6 A = (y-7)

4^

^-y^

-6

y + 14J

3y

6 nghie

m ne

n A>

<

Tuong tu coi phuon

g trin

h (2) l

-32t tre

n

3

Ta c6: f'(t) = 4t^

-3

2 = 4(t^

< =>

f (x) =

f (y+

l) <

= > x = y+ 1

3 , su

a duoc: 3y

^ -10

y +

8 = 0<:=>y= -

hoac y = 2

+ Vdi y = 2 X = 3 (thoa man)

Vay he da cho c6 nghiem (x;y) = (4; 2)

V i du 24 Giai cac he phuong trinh sau tren tap so thuc:

1

Lai gidi

x3-3

x =

(y-l)3

-7 9(

l) (1)

y-l +V

^ = , /^

i

(2)

Tu dieu tcien va

u pliuon

g trin

h (2) ta c6 x

> 1;

jy -1 > 1

Phuang trin

h (1) o x'' -3

x = (^y-1)^

3^

-

y-l , xet ha

m s

o f(t) = t'' -3t tre

n [V,+x)

Ham so dong bie

n tre

n [1

; +oo), t

a c

6 f(x) = f(^y -1) =

y =

5

Voi X = ^y-

i dug

c x = 1

; x = 2

2

\/x^Tl -3 x^

x +

2 =

0 (2)

Voi y < 0 thi VT

(I ) >

, VP(l) < 0 =>

tix phuan

g trin

h (2) cu

a h

e su

y r

a x > 2

Khid

o:

(l )c

yfj4y^

+1-1 V /

o Vx^ +

1 +

2 = 2x^y^4y^

+1 +x^y (3)

Thay 2 = x

Vx^ +

1 +

x = 2x2yj4y2

+1 +2x^y <=>

- 1+ +-

= 2yj4y

^+

1+

2y

X y

X ^

Xet ha

m so: f(t) =

0

I

Taco: f'(t) =

Vl +

t^

,=

^ = 2y c^

xy = l

Thay x

y = — va

o phuan

g trin

h (2) cu

a h

e t

a c6: x = 4 => y = —

2 8

Thu lai tha

y x = 4, y = - thoa ma

y-y + 2)

2 •

x2+y

X + N/X^+1=3>'

y + Jy^

+l

=3'' x'

y'

y-=7

x^y + 2xy^

+y

' =

9

Lai giai

1 Thay 2 = + y^ vao phuong trinh thu nhat ta dugc

2^-2>' = ( y - x ) ( x y + x^+y2)c:>2''-2>' =y3-x3 c:>2''+x^ =2>'+y3 (1) Xet ham so f(t) = 2'+ t^,t e c6 f'(t) = 2 ' l n 2 + 3t^ > 0,Vt e #, suy ra f(t) dong bien

tren M Khi do phuong trinh (1) o f(x) = f(y) <=> x = y, the vao phuong trinh thu hai ta dugc

X = y = ± 1

Vay he phuong trinh da cho c6 nghiem la (x;y) = (1;1); (-1;-1)

2 Tru ve hai phuong trinh ta dugc

x + V x ^ + l - f y + J y 2 + l l = 3>'-3'< ^ x + V x ^ + l + S " = y + J y ^ + 1 + 3 ^ ,

CO dang f(x) = f(y)

Xet ham so f(t) = t + V t ^ + 1 + 3 ' tren # Ta c6: f (t) = 1 + - ^ = ^ = + 3 ' l n 3 > 0,Vt e #

=> f(t) dong bien tren R

Khi do f(x) = f(y) <=> x = y, the vao phuong trinh thu nhat ta dugc

Suy ra g(x) dong bien tren R Boi vay g(x) = g(0) <=> x = 0

Vay he phuong trinh da cho c6 nghiem duy nhat (x;y) = (0;0)

y ( x ^ - y ^ ) = 7 y(x + y y = 9 = > x > y > 0

3 He da cho <=> T u phuong trinh thu hai suy ra x = - = - y

Thay vao phuong trinh dau ta dugc: y 3^8 - y - y ^ = 7

Dat t = ^ > 0 ta CO phuong trinh: t" = 7 o t ' - ( 3 - t ' ) ' + 7 t = 0

Xet ham so f(t) = t" - (3 - 1 ' + 7t = 0, ta c6: f '(t) = 9t' + 9t' (3 - 1 ' ) ' + 7 > 0; Vt > 0

Vay ha

m s

o f(t) don

g bie

n tre

n khoan

g (0;+co) ne

l)

Vi d

u 26 Gia

>8X +

1

3 ^log2(

2 + x) + log,(

4-Vl 8-

x)

<0

1^

11

u(4;+Go )

0 = (a' + m)' -[(

2m + 10)a' - a-20 + m''

Ta chon

m debie'u thuc tron^

dd'u phdn tick la

1 hhng d&ng thiec:

A' = l-4(

2m+10)(m 20) =

i(

3 + VT7) hoa

c

—

^x

(5-V2T)

^-3 Die

u kien:

2 + x

>0, 18-x

^O

4-^18-

x >

0 o-2<

x<

18

Ket h o p v o i dieu kien, ta c6 nghiem ciia bat p h u o n g trinh la: - 2 < x < 2

V i d u 27 Giai cac bat p h u o n g t r i n h sau tren tap so thuc:

Vay bat p h u o n g t r i n h d a cho c6 tap n g h i e m S = [3;+<»)

Vi d

u 28 Gia

1.2 X- +X +

1 2 + x -

4 ^ 4 X +

2

log3_^^^-(x^ -1) +

log3^^^(2

x -1) ^

4 Bat phuon

2 ^ ,2-7x

«>

2 x

1

, + 1

X +

4

(2 + ^/x2+l)^/x^+l

x2-3

^/(x + 4)(x

^ +

x + l)+

x +

4 (

2 + ,/x^

<0 +l x2 ^/ l) /^ ^ 2 + 4 ( x + )+ l x + 4)(x^+ ^,/(x +

o

X -

- 3 < 0 <=>

2 Die

u kie

n x > 1 Ba

^ lo8V2 i(''' -

^) ^ '°gV^2

^ ^(x

^ -1) ^ log^

i d

o x + — > 0

va ~ > 0

2x + 2 |

x2 -±

-Vay bat phuong trinh da cho c6 tap nghiem S = 3 — • 5 ;+oo

V i du 30 Giai bat phuang trinh sau tren tap so thuc: 9" + x.S" + 3 < S"""^ + 3x + 3"

Vay bat phuong trinh da cho c6 tap nghiem S = [O; 1

V i d u 31 Tim m de phuang trinh sau c6 nghiem thuc

Dat t = Phuang trinh tro thanh (t -1)^ + = m

x + 2 ] - 2 x 1 Xet ham g(x) = - ; = = = , ta c6 g'(x) = —^ , va g'(x) = O o x = -

4x^ + 1

1

Gioi han: lim g(x) =

I,

lim g(x) = -

tren (-1

; S

],iac6 f'(t

) = 2(t-l) ^

va f'(t) =

2 Gid

i han: li

Lap ban

g bie

n thien, su

y r

a phuon

g trin

h f(t) =

i v

a ch

i

khi m ^ 7

18

Vi d

u 32 Ti

m

(x 2

+l)%

0

Lot

giai

x^(-2x2+x-2) +

m~

m2)(

x2+l)%

0 (l)

o

(m-m^

)(

x^

+l)%

VX^

+1/

n M

, t

a c6: g'(x) =

va g'(x) =

0 o

x = ±

Lai c6: lim g(x)=

lim g(x) =

)^

i.

Dat t=

g(

x),

- m

^ ^ 2t^ -

1^

(3)

Xet ha

m so: f(t) = 2t^

-t' tre

n

Tac6:f'(t) = 4t-3t^ v

I

2

; 8'

UJ

: f'(t) =

0 <=>

t =

0

min f(t) =

6 nghie

m

nay xay ra k h i va chi k h i m = 16 hoac m ;t 16 va A ' = ( m +6)^ - ( m - 1 6 ) ( 3 m + 4 0 ) ^ 0

<=> 5 - 7363 < m < 5 + V363 V a y 5 - ^363 < m ^ 5 + V363 thoa m a n bai toan

-y) x

2 +

x +

y

i/x^ +2y

x +

y

Jx^

+2y + l +y + l

J.—

;x e[0

;+

oo );

lim f(x) =

0, V

x e [0; +oo) ^ f(x) don

f(0) = -

1 V

x E [0; +oo)

-y

^+

3y

3x-2 = 0

+ m = 0

CO nghie

m thuc

Dieu kien: -l<

x<

l, 0

<y

<2

-^2y-y^

+m=

0 (2)

Phuong trin

h (1)

« x

^ - 3

x = (y -1)^ - 3(y -1)

Xet ha

m s

o f(t) = t"' - 3t tren doa

n [-1

; 1] v

a ta thay ha

n f(x) = f(y-l)<=>x =

- lc:>

y =

x + l

-2V1

x^

-,; ,x

e -1;1 ,g'(x) =

Phuong trin

h (3) c

6 nghie

m x

e -1

Dteu kien du: V d i m > 1 , xet he p h u o n g t r i n h

Gia s u (xQ;yQ) la n g h i e m cua he (*)

Do d 6 m o i n g h i e m cua h e (*) d e u la n g h i e m cua he d a cho

Thay x = - 2 y v a o p h u o n g t r i n h t h u hai cua he (*) ta d u g c

He (*) CO n g h i e m , d o d o he cho c u n g c6 n g h i e m V a y m > 1 thoa m a n bai toan

BAI TAP TVr LUYEN

Bai t a p 1: G i a i cac p h u o n g t r i n h sau t r e n tap so t h u c :

x + 4->

/5-x+3x^

8x-1

-9>

0

Bai ta

p 3: Gia

i ca

c phuan

g trinh, ba

1 2 log3 (x' -

4) +

3^1og3 (

x 2)' - log3 (

x 2)' =

4

• l og jX

log2N/x

+ 2

3 log,(

x'+

4x + 6) + log,(

x'+

2)

= 1 +logg(x+ 2)

'

4

Bai ta

p 4: Gia

i phuon

g trin

h

^3x^

-x + 200

1 -\/3x^-7

x +

2002 -^

6x-200

Bai ta

p 5: Gia

Bai ta

p 6: Gia

^ +

2

il og y^

-21og2

log4(2

y +1

)2 + lo g^

7^

(9xy + 9) + (y' +3x) = 10y

9xy(y' +3x) + 3x + y' = 25y' -40

y +

16

Bai ta

p 7: Gia

6)

= y

(x2+l)

(y

-l)(x2+

6) = x(y2+l)

-x2+y2+x + y-4 = 0

'x2

4

.y2+

^ = 1

6 y x +

x^ 2

x + x^ x

-^ y

8y 3 ^3y

4 2

4 x + y

Bai ta

p 8: Gia

1 ^

x + l+\/x + 2+^

x + 3

= 0

2.

v^

+ Vx^

Bai tap 12: Cho cac so thuc a, b, c thoa m a n a > b > c > 0 Chung minh rang phuong trinh

sau C O nghiem duy nhat: V x - a - V x - b + ^ ^ = 0

V x - c